. AND DIAGRAMS OF FORCES. 201 



where Y is a function of y only. Similarly for the displacement /8 in the 

 direction of y, 



where X is a function of x only. Now the shearing stress U depends on the 

 shearing strain and the rigidity, or 



Multiplying both sides of this equation by 2(er+l) and substituting from (11), 

 (14), and (15), 



d*G d'G d'G d'G . ^ (#H d*H\ dX dY 



. dY 



Hence 



an equation which must be fulfilled by G when the body is originally without 

 strain. 



CASE II. In the second case, in which there is no strain in the direction 

 of z, we have 



Q ............................ (19). 



Substituting for R in (13), and dividing by <r+l, 



da d* r .<?(? d*G 



/OM 

 ......... (20), 



with a similar equation for . Proceeding as in the former case, we find 



dy 1 ) dx dy I a- \doc* dy*/ 



This equation is identical with that of the first case, with the exception of 



the coefficient of the part due to H, which depends on the density of the 



body, and the value of cr, the ratio of lateral expansion to longitudinal com- 

 pression. 



Hence, if the external forces are given in the two cases of no stress and 

 no strain in the direction of z, and if the density of the body or the intensity 

 VOL. ii. 26 



